B. Flag of Berland

The flag of Berland is such rectangular field n × m that satisfies following conditions:

Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
Each color should be used in exactly one stripe.

You are given a field n × m, consisting of characters 'R', 'G' and 'B'. Output "YES" (without quotes) if this field corresponds to correct flag of Berland. Otherwise, print "NO" (without quotes).

Input

The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.

Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field.

Output

Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes).

Examples

Input

6 5
RRRRR
RRRRR
BBBBB
BBBBB
GGGGG
GGGGG

Output

YES

Input

4 3
BRG
BRG
BRG
BRG

Output

YES

Input

6 7
RRRGGGG
RRRGGGG
RRRGGGG
RRRBBBB
RRRBBBB
RRRBBBB

Output

NO

Input

4 4
RRRR
RRRR
BBBB
GGGG

Output

NO

Note

The field in the third example doesn't have three parralel stripes.

Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1.

这题理解了就好做，就是条件有点复杂，需要好几次判断。（也可能是我自己想多了）

下面的代码其实可以合并条件，代码量可以缩小一半，但是我不愿改了。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <set>
  5 using namespace std;
  6 char s[105][105];
  7 set<char> st;
  8 int main(){
  9     int n,m;
 10     cin>>n>>m;
 11     bool prime=true;
 12     for(int i=1;i<=n;i++){
 13         for(int j=1;j<=m;j++){
 14             cin>>s[i][j];
 15         }
 16     }
 17     if(n%3!=0&&m%3!=0){
 18        prime= false;
 19     }else if(n%3==0&&m%3!=0){
 20         char a=s[1][1];
 21         for(int i=1;i<=n/3;i++) {
 22             for(int j=1;j<=m;j++){
 23                 if(a!=s[i][j]){
 24                     prime=false;
 25                     break;
 26                 }
 27             }if(!prime) break;
 28         }st.insert(a);
 29         char b=s[n/3+1][1];
 30         for(int i=n/3+1;i<=2*n/3;i++){
 31             for(int j=1;j<=m;j++){
 32                 if(b!=s[i][j]){
 33                     prime=false;
 34                     break;
 35                 }
 36             }if(!prime) break;
 37         }st.insert(b);
 38         char c=s[n][1];
 39         for(int i=2*n/3+1;i<=n;i++){
 40             for(int j=1;j<=m;j++){
 41                 if(c!=s[i][j]){
 42                     prime=false;
 43                     break;
 44                 }
 45             }if(!prime) break;
 46         }st.insert(c);
 47     }else if(n%3!=0&&m%3==0){
 48         char a=s[1][1];
 49         for(int i=1;i<=n;i++) {
 50             for(int j=1;j<=m/3;j++){
 51                 if(a!=s[i][j]){
 52                     prime=false;
 53                     break;
 54                 }
 55             }if(!prime) break;
 56         }st.insert(a);
 57         char b=s[1][m/3+1];
 58         for(int i=1;i<=n;i++){
 59             for(int j=m/3+1;j<=2*m/3;j++){
 60                 if(b!=s[i][j]){
 61                     prime=false;
 62                     break;
 63                 }
 64             }if(!prime) break;
 65         }st.insert(b);
 66         char c=s[1][m];
 67         for(int i=1;i<=n;i++){
 68             for(int j=2*m/3+1;j<=m;j++){
 69                 if(c!=s[i][j]){
 70                     prime=false;
 71                     break;
 72                 }
 73             }if(!prime) break;
 74         }st.insert(c);
 75     }else{
 76         if(s[1][1]==s[1][m]){
 77             char a=s[1][1];
 78             for(int i=1;i<=n/3;i++) {
 79                 for(int j=1;j<=m;j++){
 80                     if(a!=s[i][j]){
 81                         prime=false;
 82                         break;
 83                     }
 84                 }if(!prime) break;
 85             }st.insert(a);
 86             char b=s[n/3+1][1];
 87             for(int i=n/3+1;i<=2*n/3;i++){
 88                 for(int j=1;j<=m;j++){
 89                     if(b!=s[i][j]){
 90                         prime=false;
 91                         break;
 92                     }
 93                 }if(!prime) break;
 94             }st.insert(b);
 95             char c=s[n][1];
 96             for(int i=2*n/3+1;i<=n;i++){
 97                 for(int j=1;j<=m;j++){
 98                     if(c!=s[i][j]){
 99                         prime=false;
100                         break;
101                     }
102                 }if(!prime) break;
103             }st.insert(c);
104         }else{
105             char a=s[1][1];
106             for(int i=1;i<=n;i++) {
107                 for(int j=1;j<=m/3;j++){
108                     if(a!=s[i][j]){
109                         prime=false;
110                         break;
111                     }
112                 }if(!prime) break;
113             }st.insert(a);
114             char b=s[1][m/3+1];
115             for(int i=1;i<=n;i++){
116                 for(int j=m/3+1;j<=2*m/3;j++){
117                     if(b!=s[i][j]){
118                         prime=false;
119                         break;
120                     }
121                 }if(!prime) break;
122             }st.insert(b);
123             char c=s[1][m];
124             for(int i=1;i<=n;i++){
125                 for(int j=2*m/3+1;j<=m;j++){
126                     if(c!=s[i][j]){
127                         prime=false;
128                         break;
129                     }
130                 }if(!prime) break;
131             }st.insert(c);
132         }
133     }
134     if(!prime)
135         cout<<"NO"<<endl;
136     else{
137         if(st.count('R')&&st.count('B')&&st.count('G'))
138             cout<<"YES"<<endl;
139         else
140             cout<<"NO"<<endl;
141     }
142     return 0;
143 }